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[LeetCode] 2434. Using a Robot to Print the Lexicographically Smallest String_Medium tag: stack

发布时间 2023-10-12 02:33:18作者: Johnson_强生仔仔

You are given a string s and a robot that currently holds an empty string t. Apply one of the following operations until s and t are both empty:

  • Remove the first character of a string s and give it to the robot. The robot will append this character to the string t.
  • Remove the last character of a string t and give it to the robot. The robot will write this character on paper.

Return the lexicographically smallest string that can be written on the paper.

 

Example 1:

Input: s = "zza"
Output: "azz"
Explanation: Let p denote the written string.
Initially p="", s="zza", t="".
Perform first operation three times p="", s="", t="zza".
Perform second operation three times p="azz", s="", t="".

Example 2:

Input: s = "bac"
Output: "abc"
Explanation: Let p denote the written string.
Perform first operation twice p="", s="c", t="ba". 
Perform second operation twice p="ab", s="c", t="". 
Perform first operation p="ab", s="", t="c". 
Perform second operation p="abc", s="", t="".

Example 3:

Input: s = "bdda"
Output: "addb"
Explanation: Let p denote the written string.
Initially p="", s="bdda", t="".
Perform first operation four times p="", s="", t="bdda".
Perform second operation four times p="addb", s="", t="".

 

Constraints:

  • 1 <= s.length <= 105
  • s consists of only English lowercase letters.

 

这个题目的主要思路就是判断什么时候可以将t里面的last char 放进p里面，然后t实际上就是一个stack。

判断思路： 将s 利用collections.Counter() 来将每个char的frequency得到，然后得到最小的char low。只要t[-1] <= low, 即可以把t.pop()

T: O(n) S: O(n)

class Solution:
    def robotWithString(self, s: str) -> str:
        count = collections.Counter(s)
        t, p = [], ""
        low = 'a'
        for c in s:
            t.append(c)
            count[c] -= 1
            while low < 'z' and count[low] == 0:
                low = chr(ord(low) + 1)
            while t and t[-1] <= low:
                p += t.pop()
        return p